scanf_s throws exception

Question

Why does the following code throw an exception when getting to the second scanf_s after entering an number to put into the struct.

This by no means represents a complete linked list implementation.

Not sure how to get onto the next scanf_s when having entered the value? Any ideas?

EDIT: Updated code with suggested solution, but still get an AccessViolationException after first scanf_s

Code:

struct node
{
    char name[20];
    int age;
    float height;
    node *nxt;
};

int FillInLinkedList(node* temp)
{

int result;
temp = new node;

printf("Please enter name of the person");
result = scanf_s("%s", temp->name);

printf("Please enter persons age");
result = scanf_s("%d", &temp->age); // Exception here...

printf("Please enter persons height");
result = scanf_s("%f", &temp->height);

temp->nxt = NULL;
if (result >0)
    return  1;
 else return 0;
}

// calling code

int main(array<System::String ^> ^args)
{
  node temp;

  FillInLinkedList(&temp);

...

You are using managed code, why are you messing with scanf()? Use Console::ReadLine(), int::TryParse() and List. — Hans Passant, Feb 28 '10 at 12:07

shf301 · Answer 1 · 2010-02-28T13:06:39.097

5

You are using scanf_s with incorrect parameters. Take a look at the examples in the MSDN documentation for the function. It requires that you pass in the size of the buffer after the buffer for all string or character parameters. So

result = scanf_s("%s", temp->name);

should be:

 result = scanf_s("%s", temp->name, 20);

The first call to scanf_s is reading garbage off the stack because it is looking for another parameter and possibly corrupting memory.

There is no compiler error because scanf_s uses a variable argument list - the function doesn't have a fixed number of parameters so the compiler has no idea what scanf_s is expecting.

edited Feb 28 '10 at 13:06

answered Feb 28 '10 at 12:59

shf301

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A horrible decision from microsoft. – this Dec 29 '13 at 23:57
it is also good practice to include a width specifier on the %s. scanf_s("%19s", temp->name, 20); – curlyhairedgenius Nov 07 '17 at 19:29

score 3 · Accepted Answer · answered Feb 28 '10 at 11:14

3

You need

result = scanf_s("%d", &temp->age);

and

result = scanf_s("%f", &temp->height);

Reason is that sscanf (and friends) requires a pointer to the output variable so it can store the result there.

BTW, you have a similar problem with the parameter temp of your function. Since you're changing the pointer (and not just the contents of what it points to), you need to pass a double pointer so that the changes will be visible outside your function:

int FillInLinkedList(node** temp)

And then of course you'll have to make the necessary changes inside the function.

answered Feb 28 '10 at 11:14

Martin B

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@Martin: Funny you mention pointer to pointers, that's the whole reason I was attempting to write this, is to learn about pointers to pointers, seems I haven't fully got the idea yet... – Tony The Lion Feb 28 '10 at 11:19
@Tony: Take a look at this question -- the highest voted answer has quite a good description: http://stackoverflow.com/questions/897366/how-do-pointer-to-pointers-work-in-c – Martin B Feb 28 '10 at 16:30

score 2 · Answer 3 · answered Feb 28 '10 at 11:16

scanf() stores data into variables, so you need to pass the address of the variable (or its pointer)
Example:

char string[10];
int n;
scanf("%s", string); //string actually points to address of
                     //first element of string array
scanf("%d", &n); // &n is the address of the variable 'n'

score 1 · Answer 4 · answered Feb 28 '10 at 11:13

1

%19c should be %s
temp->age should be &temp-age
temp->height should be &temp->height
Your compiler should be warning you about these errors

answered Feb 28 '10 at 11:13

Paul R

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Did you make all 3 of the above changes ? – Paul R Feb 28 '10 at 11:19
When you test are you entering less than 20 characters for the name ? – Paul R Feb 28 '10 at 12:03

score 1 · Answer 5 · answered Feb 28 '10 at 11:13

1

I believe you need to pass parameters to scanf() functions by address. i.e. &temp->age

otherwise temp-age will be interpreted as a pointer, which will most likely crash your program.

answered Feb 28 '10 at 11:13

Ferruccio

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scanf_s throws exception

5 Answers5